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In mathematics, specifically the theory of Lie algebras, **Lie's theorem** states that,^{[1]} over an algebraically closed field of characteristic zero, if is a finite-dimensional representation of a solvable Lie algebra, then there's a flag of invariant subspaces of with , meaning that for each and *i*.

Put in another way, the theorem says there is a basis for *V* such that all linear transformations in are represented by upper triangular matrices.^{[2]} This is a generalization of the result of Frobenius that commuting matrices are simultaneously upper triangularizable, as commuting matrices generate an abelian Lie algebra, which is a fortiori solvable.

A consequence of Lie's theorem is that any finite dimensional solvable Lie algebra over a field of characteristic 0 has a nilpotent derived algebra (see #Consequences). Also, to each flag in a finite-dimensional vector space *V*, there correspond a Borel subalgebra (that consist of linear transformations stabilizing the flag); thus, the theorem says that is contained in some Borel subalgebra of .^{[1]}

For algebraically closed fields of characteristic *p*>0 Lie's theorem holds provided the dimension of the representation is less than *p* (see the proof below), but can fail for representations of dimension *p*. An example is given by the 3-dimensional nilpotent Lie algebra spanned by 1, *x*, and *d*/*dx* acting on the *p*-dimensional vector space *k*[*x*]/(*x*^{p}), which has no eigenvectors. Taking the semidirect product of this 3-dimensional Lie algebra by the *p*-dimensional representation (considered as an abelian Lie algebra) gives a solvable Lie algebra whose derived algebra is not nilpotent.

The proof is by induction on the dimension of and consists of several steps. (Note: the structure of the proof is very similar to that for Engel's theorem.) The basic case is trivial and we assume the dimension of is positive. We also assume *V* is not zero. For simplicity, we write .

**Step 1**: Observe that the theorem is equivalent to the statement:^{[3]}

- There exists a vector in
*V*that is an eigenvector for each linear transformation in .

Indeed, the theorem says in particular that a nonzero vector spanning is a common eigenvector for all the linear transformations in . Conversely, if *v* is a common eigenvector, take to its span and then admits a common eigenvector in the quotient ; repeat the argument.

**Step 2**: Find an ideal of codimension one in .

Let be the derived algebra. Since is solvable and has positive dimension, and so the quotient is a nonzero abelian Lie algebra, which certainly contains an ideal of codimension one and by the ideal correspondence, it corresponds to an ideal of codimension one in .

**Step 3**: There exists some linear functional in such that

is nonzero. This follows from the inductive hypothesis (it is easy to check that the eigenvalues determine a linear functional).

**Step 4**: is a -invariant subspace. (Note this step proves a general fact and does not involve solvability.)

Let , , then we need to prove . If then it's obvious, so assume and set recursively . Let and be the largest such that are linearly independent. Then we'll prove that they generate *U* and thus is a basis of *U*. Indeed, assume by contradiction that it's not the case and let be the smallest such that , then obviously . Since are linearly dependent, is a linear combination of . Applying the map it follows that is a linear combination of . Since by the minimality of *m* each of these vectors is a linear combination of , so is , and we get the desired contradiction. We'll prove by induction that for every and there exist elements of the base field such that and

The case is straightforward since . Now assume that we have proved the claim for some and all elements of and let . Since is an ideal, it's , and thus

and the induction step follows. This implies that for every the subspace *U* is an invariant subspace of *X* and the matrix of the restricted map in the basis is upper triangular with diagonal elements equal to , hence . Applying this with instead of *X* gives . On the other hand, *U* is also obviously an invariant subspace of *Y*, and so

since commutators have zero trace, and thus . Since is invertible (because of the assumption on the characteristic of the base field), and

and so .

**Step 5**: Finish up the proof by finding a common eigenvector.

Write where *L* is a one-dimensional vector subspace. Since the base field is algebraically closed, there exists an eigenvector in for some (thus every) nonzero element of *L*. Since that vector is also eigenvector for each element of , the proof is complete.

The theorem applies in particular to the adjoint representation of a (finite-dimensional) solvable Lie algebra over an algebraically closed field of characteristic zero; thus, one can choose a basis on with respect to which consists of upper triangular matrices. It follows easily that for each , has diagonal consisting of zeros; i.e., is a strictly upper triangular matrix. This implies that is a nilpotent Lie algebra. Moreover, if the base field is not algebraically closed then solvability and nilpotency of a Lie algebra is unaffected by extending the base field to its algebraic closure. Hence, one concludes the statement (the other implication is obvious):^{[4]}

*A finite-dimensional Lie algebra over a field of characteristic zero is solvable if and only if the derived algebra is nilpotent.*

Lie's theorem also establishes one direction in Cartan's criterion for solvability:

*If*V*is a finite-dimensional vector space over a field of characteristic zero and a Lie subalgebra, then is solvable if and only if for every and .*^{[5]}

Indeed, as above, after extending the base field, the implication is seen easily. (The converse is more difficult to prove.)

Lie's theorem (for various *V*) is equivalent to the statement:^{[6]}

*For a solvable Lie algebra over an algebraically closed field of characteristic zero, each finite-dimensional simple -module (i.e., irreducible as a representation) has dimension one.*

Indeed, Lie's theorem clearly implies this statement. Conversely, assume the statement is true. Given a finite-dimensional -module *V*, let be a maximal -submodule (which exists by finiteness of the dimension). Then, by maximality, is simple; thus, is one-dimensional. The induction now finishes the proof.

The statement says in particular that a finite-dimensional simple module over an abelian Lie algebra is one-dimensional; this fact remains true over any base field since in this case every vector subspace is a Lie subalgebra.^{[7]}

Here is another quite useful application:^{[8]}

*Let be a finite-dimensional Lie algebra over an algebraically closed field of characteristic zero with radical . Then each finite-dimensional simple representation is the tensor product of a simple representation of with a one-dimensional representation of (i.e., a linear functional vanishing on Lie brackets).*

By Lie's theorem, we can find a linear functional of so that there is the weight space of . By Step 4 of the proof of Lie's theorem, is also a -module; so . In particular, for each , . Extend to a linear functional on that vanishes on ; is then a one-dimensional representation of . Now, . Since coincides with on , we have that is trivial on and thus is the restriction of a (simple) representation of .

- Engel's theorem, which concerns a nilpotent Lie algebra.
- Lie–Kolchin theorem, which is about a (connected) solvable linear algebraic group.

- ^
^{a}^{b}Serre 2001, Theorem 3 **^**Humphreys 1972, Ch. II, § 4.1., Corollary A.**^**Serre 2001, Theorem 3″**^**Humphreys 1972, Ch. II, § 4.1., Corollary C.**^**Serre 2001, Theorem 4**^**Serre 2001, Theorem 3'**^**Jacobson 1979, Ch. II, § 6, Lemma 5.**^**Fulton & Harris 1991, Proposition 9.17.

- Fulton, William; Harris, Joe (1991).
*Representation theory. A first course*. Graduate Texts in Mathematics, Readings in Mathematics. Vol. 129. New York: Springer-Verlag. doi:10.1007/978-1-4612-0979-9. ISBN 978-0-387-97495-8. MR 1153249. OCLC 246650103. - Humphreys, James E. (1972),
*Introduction to Lie Algebras and Representation Theory*, Berlin, New York: Springer-Verlag, ISBN 978-0-387-90053-7. - Jacobson, Nathan (1979),
*Lie algebras*(Republication of the 1962 original ed.), New York: Dover Publications, Inc., ISBN 0-486-63832-4, MR 0559927 - Serre, Jean-Pierre (2001),
*Complex Semisimple Lie Algebras*, Berlin: Springer, doi:10.1007/978-3-642-56884-8, ISBN 3-5406-7827-1, MR 1808366